What is the extraneous solution to these equations? $\dfrac{x^2 + 9x}{x + 1} = \dfrac{-x - 9}{x + 1}$
Explanation: Multiply both sides by $x + 1$ $ \dfrac{x^2 + 9x}{x + 1} (x + 1) = \dfrac{-x - 9}{x + 1} (x + 1)$ $ x^2 + 9x = -x - 9$ Subtract $-x - 9$ from both sides: $ x^2 + 9x - (-x - 9) = -x - 9 - (-x - 9)$ $ x^2 + 9x + x + 9 = 0$ $ x^2 + 10x + 9 = 0$ Factor the expression: $ (x + 9)(x + 1) = 0$ Therefore $x = -9$ or $x = -1$ At $x = -1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -1$, it is an extraneous solution.